Question: Solve for $x$, ignoring any extraneous solutions: $\dfrac{x^2 + 4x}{x + 2} = \dfrac{-8x - 20}{x + 2}$
Solution: Multiply both sides by $x + 2$ $ \dfrac{x^2 + 4x}{x + 2} (x + 2) = \dfrac{-8x - 20}{x + 2} (x + 2)$ $ x^2 + 4x = -8x - 20$ Subtract $-8x - 20$ from both sides: $ x^2 + 4x - (-8x - 20) = -8x - 20 - (-8x - 20)$ $ x^2 + 4x + 8x + 20 = 0$ $ x^2 + 12x + 20 = 0$ Factor the expression: $ (x + 2)(x + 10) = 0$ Therefore $x = -2$ or $x = -10$ However, the original expression is undefined when $x = -2$. Therefore, the only solution is $x = -10$.